Let $V$ an inner product space, $T$ is a projection operator on $V$ iff $T^2=T$, which is the definition from Wiki. And I also know a result from my textbook that $T$ is an orthogonal projection iff $T^*$ exist and $T^2=T=T^*.$ I thought this result should be a more clean definition of orthogonal projection, but in the book the definition is $T$ is orthogonal projection if $R(T)^\perp=N(T)$ and $N(T)^\perp=R(T),$ and when the author proving that two orthogonal projection are the same(uniqueness) he says "since every projection is uniquely determined by its range and null space, ...". From this claim can I say that every projection is uniquely determined by a direct sum $V=W_1\oplus W_2$ of $V$, would this equivalent to $T^2=T$? My idea is that since $V=W_1\oplus W_2$ then given any $x\in V, x=x_1+x_2,$ where $x_1\in W_1,x_2\in W_2$ will be a unique decomposition of $x$ thus there is no ambiguity on what should be the image under the projection.
Book I'm reading: Linear Algebra, 4th, Friedberg, p.398 definition, p.399 4th line is the claim
You can always write $V=N(T)\oplus N(T)^\perp$, and by the definition of the orthogonal operator, $V=N(T)\oplus R(T)$. Conversely, if $V=W_1\oplus W_2$, any element $v\in V$ can be written in the form of $v=w_1+w_2$ where $w_1\in W_1, w_2\in W_2$ are uniquely determined. Now define an operator $T$ to be $Tv=w_1$. Check that this operator is well-defined, linear, and is a projection, and $N(T)=W_2$ and $R(T)=W_1$.